Yes, (g x) is the second argument to f. Consider the type signature:

(<*>) :: Applicative f => f (a -> b) -> f a -> f b

In this case, the type of f is ((->) r). Specialized to that type:

(<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)

f <*> g = \x -> f x (g x)

Breaking down the pieces...

f :: r -> a -> b

g :: r -> a

x :: r

(g x) :: a

(f x (g x)) :: b

The example is made a bit confusing by tossing in an fmap. As far as the definition above is concerned, 'f' in the example is ((+) <$> (+3)) and that has to be resolved before looking at <*>.

On Mon, Aug 22, 2016 at 9:07 AM, Olumide <50295@web.de> wrote:

Hi List,

I'm struggling to relate the definition of a function as a function

instance Applicative ((->) r) where
    pure x = (\_ -> x)
    f <*> g = \x -> f x (g x)

with the following expression

ghci> :t (+) <$> (+3) <*> (*100)
(+) <$> (+3) <*> (*100) :: (Num a) => a -> a
ghci> (+) <$> (+3) <*> (*100) $ 5
508

>From chapter 11 of LYH http://goo.gl/7kl2TM .

I understand the explanation in the book: "we're making a function that will use + on the results of (+3) and (*100) and return that. To demonstrate on a real example, when we did (+) <$> (+3) <*> (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in 8 and 500. Then, + gets called with 8 and 500, resulting in 508."

The problem is that I can't relate that explanation with the definition of a function as an applicative; especially f <*> g = \x -> f x (g x) . Is (g x) the second argument to f?

Regards,

- Olumide
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